What does the force of gravity depend on?

On Earth, the gravitational Field strength is much than on the moon. If a piece of rock was taken from the Moon to the Earth, st

nikklg [1K]

Ok so I’m pretty sure the answer would be 2 because the mass of the rock would have the same mass on Earth as it has on the moon. Also the Density of a solid object remains constant meaning it doesn’t change. But the weight would change because the Earths gravitational pull is more than that of the moon. I hope this helped!

7 0

3 years ago

Explain how high-pressure and low pressure systems are different

Studentka2010 [4]

Low pressure has a bit less of a function than high pressure, high pressure is more useful in certain terms

5 0

2 years ago

100 POINTS FOR CORRECT ANSWER/EXPLANATION

Shalnov [3]

Answer:

6 N

Explanation:

Let's start with the small block m on top. There are four forces:

Weight force mg pulling down, normal force N₁ pushing up, tension force T pulling right, and friction force N₁μ pushing left (opposing the direction of motion).

Now let's look at the large block M on bottom. There are seven forces:

Normal force N₁ pushing down (opposite and equal from block m),

Friction force N₁μ pushing right (opposite and equal from block m),

Weight force Mg pulling down,

Tension force T pulling right,

Applied force F pulling left,

Normal force N₂ pushing up,

and friction force N₂μ pushing right (opposing the direction of motion).

So you've correctly identified the free body diagrams.

Now apply Newton's second law. Sum of forces in the y direction for block m:

∑F = ma

N₁ − mg = 0

N₁ = mg

Sum of forces in the x direction:

∑F = ma

T − N₁μ = 0

T = N₁μ

T = mgμ

Sum of forces in the y direction for block M:

∑F = ma

-N₁ − Mg + N₂ = 0

N₂ = N₁ + Mg

N₂ = mg + Mg

Sum of forces in the x direction:

∑F = ma

N₁μ + T − F + N₂μ = 0

F = N₁μ + T + N₂μ

F = mgμ + mgμ + (mg + Mg)μ

F = gμ(3m + M)

Since M = 2m:

F = 5gμm

Plug in values:

F = 5 (10 m/s²) (0.400) (0.300 kg)

F = 6 N

8 0

3 years ago

Read 2 more answers

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago

Jane starts from her house to take a stroll in her neighborhood. After walking for 2 hours at a steady pace, she has walked 4 mi

9966 [12]

Answer:

C. 2 mph

Explanation:

speed equals total distance covered in total time of of travel

since,

total distance = 4 miles

time taken = 2 hrs

speed = 4m / 2hrs =2mph

3 0

3 years ago

Read 2 more answers